MHB Deriving a Vector Expression from Given Direction

  • Thread starter Thread starter Drain Brain
  • Start date Start date
  • Tags Tags
    Expression Vector
Click For Summary
To derive a vector expression from the direction defined by dx=2dy=-2dz, one must first express the relationships between the variables. This can be done by setting a parameter, such as t, to represent the changes in x, y, and z based on the given ratios. The directional derivative of a scalar function can then be calculated using the gradient and the unit vector derived from these relationships. Understanding how to convert the directional ratios into a vector representation is crucial for solving the problem. Seeking clarification on these steps is essential for progressing in the calculation of the directional derivative.
Drain Brain
Messages
143
Reaction score
0
How can you derive an expression from the given direction defined by dx=2dy=-2dz. This is just a part of a problem that I'am solving. I have no idea how to get a vector expression using the given info. please help. Thanks!

by the way I'm trying to find the directional derivative of a certain scalar function in the direction given above. By I don't know how to get a vector representayion of the given direction. Please help me!
 
Last edited:
Physics news on Phys.org
is that possible? use derivative.
 
What? I Know that I would use partial derivative here. But what I need help with is finding the vector representation of the given direction. Please anybody knows how to do it tell me. Thanks.!
 
Hey guys, do you have any idea on how to do this? I want to know it so badly. Please give me an insight.
 
Thread 'Problem with calculating projections of curl using rotation of contour'
Hello! I tried to calculate projections of curl using rotation of coordinate system but I encountered with following problem. Given: ##rot_xA=\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}=0## ##rot_yA=\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}=1## ##rot_zA=\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=0## I rotated ##yz##-plane of this coordinate system by an angle ##45## degrees about ##x##-axis and used rotation matrix to...

Similar threads

Replies
8
Views
3K
Replies
2
Views
4K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
955
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K